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| 型号： | LT1111IS8#TRPBF |
| 厂商： | Linear Technology |
| 文件页数： | 8/16页 |
| 文件大小： | 0K |
| 描述： | IC REG BUCK BOOST INV ADJ 8SOIC |
| 标准包装： | 2,500 |
| 类型： | 降压（降压），升压（升压），反相 |
| 输出类型： | 可调式 |
| 输出数： | 1 |
| 输出电压： | 可调 |
| 输入电压： | 2 V ~ 30 V |
| 频率 - 开关： | 72kHz |
| 电流 - 输出： | 400mA |
| 同步整流器： | 无 |
| 工作温度： | -40°C ~ 105°C |
| 安装类型： | 表面贴装 |
| 封装/外壳： | 8-SOIC（0.154"，3.90mm 宽） |
| 包装： | 带卷 (TR) |
| 供应商设备封装： | 8-SOIC |
LT1111
A PPLICATI
S I FOR ATIO
Picking an inductor value of 47 μ H with 0.2 ? DCR results
in a peak switch current of:
V D = diode drop (0.5V for a 1N5818)
I OUT = output current
1 . 0 ? ?
? = 623 mA .
I PEAK
=
?
4 .5V ?
?
1 – e
–1.0 ? ? 7 μ s ?
47 μ H
?
?
( 8 )
V OUT = output voltage
V IN = minimum input voltage
V SW is actually a function of switch current which is in turn
a function of V IN , L, time, and V OUT . To simplify, 1.5V can
( )( ) 2 = 9 . 1 μ J
E L =
47 μ H 0 . 623 A
V IN MIN ? V SW ? V OUT
Substituting I PEAK into Equation 4 results in:
1
( 9 )
2
Since 9.1 μ J > 6.7 μ J, the 47 μ H inductor will work. This
trial-and-error approach can be used to select the opti-
mum inductor. Keep in mind the switch current maximum
rating of 1.5A. If the calculated peak current exceeds this,
consider using the LT1110. The 70% duty cycle of the
LT1110 allows more energy per cycle to be stored in the
inductor, resulting in more output power.
A resistor can be added in series with the I LIM pin to invoke
switch current limit. The resistor should be picked so the
be used for V SW as a very conservative value.
Once I PEAK is known, inductor value can be derived from:
)
L = ? t ON ( 11
I PEAK
where t ON = switch-on time (7 μ s).
Next, the current limit resistor R LIM is selected to give
I PEAK from the R LIM Step-Down Mode curve. The addition
of this resistor keeps maximum switch current constant as
the input voltage is increased.
As an example, suppose 5V at 300mA is to be generated
from a 12V to 24V input. Recalling Equation (10),
2 ( 300 mA ) ? 5 + 0 . 5 ?
? ? 12 – 1 . 5 + 0 . 5 ? ?
calculatedI PEAK atminimumV IN isequaltotheMaximum
Switch Current (from Typical Performance Characteristic
curves). Then, as V IN increases, switch current is held
constant, resulting in increasing efficiency.
I PEAK =
0 . 50
? ? = 600 mA
( 12 )
Inductor Selection — Step-Down Converter
Next, inductor value is calculated using Equation (11):
L =
7 μ s = 64 μ H .
Thestep-downcase(Figure5)differsfromthestep-upin
that the inductor current flows through the load during
both the charge and discharge periods of the inductor.
Current through the switch should be limited to ~650mA
12 – 1.5 – 5
600 mA
Use the next lowest standard value (56 μ H).
( 13 )
in this mode. Higher current can be obtained by using an
external switch (see Figure 6). The I LIM pin is the key to
successful operation over varying inputs.
After establishing output voltage, output current and input
voltage range, peak switch current can be calculated by the
formula:
Then pick R LIM from the curve. For I PEAK = 600mA, R LIM
= 56 ? .
Inductor Selection — Positive-to-Negative Converter
Figure 7 shows hookup for positive-to-negative conver-
sion. All of the output power must come from the inductor.
? ? V IN – V SW + V D ? ?
I PEAK =
2 I OUT
DC
? V OUT + V D ?
? ?
( 10 )
In this case,
P L = ( ? V OUT ? + V D )( I OUT )
(14)
where DC = duty cycle (0.50)
V SW = switch drop in step-down mode
In this mode the switch is arranged in common collector
or step-down mode. The switch drop can be modeled as
a 0.75V source in series with a 0.65 ? resistor. When the
1111fd
8
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相关代理商/技术参数 |
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